This article is a continuation of "teaching rolling motion. A description of the notation that use another article "Teaching Dynamics rotation" can be found. In article rolling motion, I analyzed the motion of a sphere rolling down a slope. I guess no slippage and then determined how State hypothesis imposed on the coefficient of friction. This article describes a situation for which this condition is not satisfied - landslide occurs.
Problem. A sphere rolls down a steep enough that slide occurs. What is linear acceleration of Center of mass of the sphere? What is the angular acceleration of sphere around the axis through its centre of mass? The mass of the sphere is M, its RADIUS is R, his moment of inertia about the center of mass is Icm = 2(MR**2)/5, tilt angle is th and the coefficient of kinetic friction between the sphere and the slope is the United Kingdom.
Analysis. We will use a system of inertial reference with an x-axis pointing to the bottom of the slope and an axis is perpendicular to the slope. Slope exerts a normal force n and friction force f on the sphere. The weight of the sphere has components MGsin (th) slope and MGcos (th) perpendicular to the slope. With the help of a free body diagram, we apply the equations of motion:
..................................The second law of Newton...Rotation equation of motion
...................Sum (FX) = MAx....Sum (FY) = may...Sum (Text) = IcmA
............MGsin (th) - f = MAx....N - MGcos (th) = 0... fR = IcmA
From hoop slips, f = UkN. Leave the equation y, N = MGcos (th). By combining these two, we find that f = UkMGcos (th). When it is overridden in the equation of x, we
...............................................MGsin (th) - UkMGcos (th) = Max;
so...............................................AX = G (sin (th) - Ukcos (th)).
Note that there is rotation equation to determine the acceleration. Also note that it is the acceleration would a box down the slope. For each of them force net downward slope is Mgsin (th) and the net force up the slope is UkMGcos (th). But we need the equation rotation movement in order to determine the angular acceleration; It is
........................(A = fR/Icm = (UkMGcos (th)) R /(2MR**2)/5) = 5UkGcos (th) / 2R.
We see slides case causes a simpler solution to problem that no slippage. However, even for drag the condition imposed on the coefficient of friction must be determined with the problem of the non - slip.
Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.
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